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{"segment":[{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Ch\u1ea5t t\u00e1c d\u1ee5ng v\u1edbi n\u01b0\u1edbc t\u1ea1o ra dung d\u1ecbch baz\u01a1 l\u00e0:","select":["A. $CO_2$. ","B. $Na_2O$.","C. $SO_2$ ","D. $P_2O_5$"],"hint":"","explain":"<span class='basic_left'>Oxit baz\u01a1 c\u1ee7a kim lo\u1ea1i ki\u1ec1m, ki\u1ec1m th\u1ed5 t\u00e1c d\u1ee5ng v\u1edbi n\u01b0\u1edbc t\u1ea1o th\u00e0nh dung d\u1ecbch baz\u01a1. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n B th\u1ecfa m\u00e3n<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1831},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" L\u01b0u hu\u1ef3nh trioxit $(SO_3)$ t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi:","select":["A. N\u01b0\u1edbc, s\u1ea3n ph\u1ea9m l\u00e0 baz\u01a1.","B. Axit, s\u1ea3n ph\u1ea9m l\u00e0 baz\u01a1.","C. N\u01b0\u1edbc, s\u1ea3n ph\u1ea9m l\u00e0 axit","D. Baz\u01a1, s\u1ea3n ph\u1ea9m l\u00e0 axit."],"hint":"","explain":"<span class='basic_left'>L\u01b0u hu\u1ef3nh trioxit l\u00e0 oxit axit c\u00f3 ph\u1ea3n \u1ee9ng v\u1edbi n\u01b0\u1edbc t\u1ea1o axit sunfuric t\u01b0\u01a1ng \u1ee9ng. T\u00e1c d\u1ee5ng v\u1edbi baz\u01a1 s\u1ea3n ph\u1ea9m la mu\u1ed1i v\u00e0 n\u01b0\u1edbc. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n C th\u1ecfa m\u00e3n <br\/>$S{{O}_{3}}+{{H}_{2}}O\\to {{H}_{2}}S{{O}_{4}}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":1}]}],"id_ques":1832},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Hai oxit t\u00e1c d\u1ee5ng v\u1edbi nhau t\u1ea1o th\u00e0nh mu\u1ed1i l\u00e0:","select":["A. $CO_2$ v\u00e0 $BaO$.","B. $K_2O$ v\u00e0 $NO$.","C. $Fe_2O_3$ v\u00e0 $SO_3$. ","D. $MgO$ v\u00e0 $CO$."],"hint":"","explain":"<span class='basic_left'>Oxit axit t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi oxit baz\u01a1 c\u1ee7a kim lo\u1ea1i ki\u1ec1m, ki\u1ec1m th\u1ed5 t\u1ea1o mu\u1ed1i. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n A th\u1ecfa m\u00e3n <br\/>$BaO+C{{O}_{2}}\\to BaC{{O}_{3}}\\downarrow $ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1833},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" C\u1eb7p ch\u1ea5t t\u00e1c d\u1ee5ng v\u1edbi nhau t\u1ea1o th\u00e0nh mu\u1ed1i v\u00e0 n\u01b0\u1edbc:","select":["A. Magie v\u00e0 dung d\u1ecbch axit sunfuric ","B. Magie oxit v\u00e0 dung d\u1ecbch axit sunfuric","C. Magie nitrat v\u00e0 natri hidroxit ","D. Magie clorua v\u00e0 natri clorua"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& Mg+{{H}_{2}}S{{O}_{4}}\\to MgS{{O}_{4}}+{{H}_{2}} \\\\ & MgO+{{H}_{2}}S{{O}_{4}}\\to MgS{{O}_{4}}+{{H}_{2}}O \\\\ & Mg{{(N{{O}_{3}})}_{2}}+NaOH\\to Mg{{(OH)}_{2}}\\downarrow +2NaN{{O}_{3}} \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":1834},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" D\u00e3y ch\u1ea5t g\u1ed3m c\u00e1c oxit baz\u01a1:","select":["A. $CuO, NO, MgO, CaO.$ ","B. $CuO, CaO, MgO, Na_2O.$","C. $CaO, CO_2, K_2O, Na_2O.$","D. $K_2O, FeO, P_2O_5, Mn_2O_7.$"],"hint":"","explain":"<span class='basic_left'>Oxit baz\u01a1 l\u00e0 oxit c\u1ee7a kim lo\u1ea1i. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n B th\u1ecfa m\u00e3n<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":1835},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" $MgCO_3$ t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch $HCl$ sinh ra:","select":["A. Ch\u1ea5t kh\u00ed ch\u00e1y \u0111\u01b0\u1ee3c trong kh\u00f4ng kh\u00ed ","B. Ch\u1ea5t kh\u00ed l\u00e0m v\u1ea9n \u0111\u1ee5c n\u01b0\u1edbc v\u00f4i trong","C. Ch\u1ea5t kh\u00ed duy tr\u00ec s\u1ef1 ch\u00e1y v\u00e0 s\u1ef1 s\u1ed1ng ","D. Ch\u1ea5t kh\u00ed kh\u00f4ng tan trong n\u01b0\u1edbc"],"hint":"","explain":"<span class='basic_left'>$MgC{{O}_{3}}+2HCl\\to MgC{{l}_{2}}+C{{O}_{2}}\\uparrow +{{H}_{2}}O$ <br\/> $CO_2$ l\u00e0 ch\u1ea5t kh\u00ed kh\u00f4ng duy tr\u00ec s\u1ef1 ch\u00e1y, l\u00e0m \u0111\u1ee5c n\u01b0\u1edbc v\u00f4i trong <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":1}]}],"id_ques":1836},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" D\u00e3y oxit v\u1eeba t\u00e1c d\u1ee5ng n\u01b0\u1edbc, v\u1eeba t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch ki\u1ec1m l\u00e0:","select":["A. $CuO, Fe_2O_3, SO_2, CO_2.$","B. $CaO, CuO, CO, N_2O_5.$","C. $SO_2, MgO, CuO, Ag_2O.$ ","D. $CO_2, SO_2, P_2O_5, SO_3.$"],"hint":"","explain":"<span class='basic_left'>C\u00e1c oxit axit v\u1eeba t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi n\u01b0\u1edbc, v\u1eeba t\u00e1c d\u1ee5ng \u0111\u01b0\u1ee3c v\u1edbi dung d\u1ecbch ki\u1ec1m. Ch\u1ec9 c\u00f3 \u0111\u00e1p \u00e1n D th\u1ecfa m\u00e3n<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":1}]}],"id_ques":1837},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" D\u00e3y ch\u1ea5t t\u00e1c d\u1ee5ng v\u1edbi dung d\u1ecbch $H_2SO_4$ lo\u00e3ng t\u1ea1o th\u00e0nh dung d\u1ecbch c\u00f3 m\u00e0u xanh lam:","select":["A. $CuO, MgCO_3$ ","B. $Cu, CuO$","C. $Cu(NO_3)_2, Cu$ ","D. $CuO, Cu(OH)_2$"],"hint":"","explain":"<span class='basic_left'>Mu\u1ed1i c\u00f3 m\u00e0u xanh lam l\u00e0 c\u00e1c mu\u1ed1i c\u1ee7a Cu. N\u00ean ch\u1ea5t t\u00e1c d\u1ee5ng ch\u1ec9 c\u00f3 th\u1ec3 l\u00e0 oxit v\u00e0 hi\u0111roxit c\u1ee7a Cu<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":1838},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" H\u00f2a tan 16 gam $SO_3$ trong n\u01b0\u1edbc thu \u0111\u01b0\u1ee3c 250 ml dung d\u1ecbch. N\u1ed3ng \u0111\u1ed9 mol dung d\u1ecbch axit thu \u0111\u01b0\u1ee3c l\u00e0:","select":["A. 0,2 M ","B. 0,4 M","C. 0,6 M ","D. 0,8 M"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{S{{O}_{3}}}}=\\dfrac{16}{80}=0,2\\,mol \\\\ & S{{O}_{3}}+{{H}_{2}}O\\to {{H}_{2}}S{{O}_{4}} \\\\ & {{n}_{{{H}_{2}}S{{O}_{4}}}}={{n}_{S{{O}_{3}}}}=0,2\\,mol \\\\ & {{C}_{M\\,({{H}_{2}}S{{O}_{4}})}}=\\dfrac{0,2}{0,25}=0,8\\,M \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":1839},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Dung d\u1ecbch \u0111\u01b0\u1ee3c t\u1ea1o th\u00e0nh t\u1eeb l\u01b0u hu\u1ef3nh \u0111ioxit v\u1edbi n\u01b0\u1edbc c\u00f3 :","select":["A. pH = 7 ","B. pH > 7","C. pH < 7 ","D. pH = 8"],"hint":"","explain":"<span class='basic_left'>$S{{O}_{2}}+{{H}_{2}}O\\to {{H}_{2}}S{{O}_{3}}$ <br\/>Dung d\u1ecbch H2SO3 c\u00f3 m\u00f4i tr\u01b0\u1eddng axit, pH < 7<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1840},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" Cho 10,5 gam h\u1ed7n h\u1ee3p hai kim lo\u1ea1i Zn, Cu v\u00e0o dung d\u1ecbch $H_2SO_4$ lo\u00e3ng d\u01b0, ng\u01b0\u1eddi ta thu \u0111\u01b0\u1ee3c 2,24 l\u00edt kh\u00ed (\u0111ktc). Ph\u1ea7n tr\u0103m theo kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a kim lo\u1ea1i Zn trong h\u1ed7n h\u1ee3p ban \u0111\u1ea7u l\u00e0","select":["A. 61,9% ","B. 63% ","C. 61,5% ","D. 65% "],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{{{H}_{2}}}}=\\dfrac{2,24}{22,4}=0,1\\,mol \\\\ & Zn+{{H}_{2}}S{{O}_{4}}\\to ZnS{{O}_{4}}+{{H}_{2}} \\\\ & {{n}_{Zn}}={{n}_{{{H}_{2}}}}=0,1\\,mol \\\\ & {{m}_{Zn}}=0,1.65=6,5\\,gam \\\\ & \\%{{m}_{Zn}}=\\dfrac{6,5}{10,5}.100\\%=61,9\\% \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1841},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" \u0110\u1ec3 trung h\u00f2a 112 gam dung d\u1ecbch KOH 50% th\u00ec c\u1ea7n d\u00f9ng bao nhi\u00eau gam dung d\u1ecbch axit sunfuric 9,8%:","select":["A. 400 g ","B. 420 g","C. 500 g ","D. 570 g"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{m}_{KOH}}=\\dfrac{112.50\\%}{100\\%}=56\\,gam \\\\ & {{n}_{KOH}}=\\dfrac{56}{56}=1\\,mol \\\\ & 2KOH+{{H}_{2}}S{{O}_{4}}\\to {{K}_{2}}S{{O}_{4}}+2{{H}_{2}}O \\\\ & {{n}_{{{H}_{2}}S{{O}_{4}}}}=\\dfrac{1}{2}{{n}_{KOH}}=0,5\\,mol \\\\ & {{m}_{{{H}_{2}}S{{O}_{4}}}}=0,5.98=49\\,gam \\\\ & {{m}_{\\text{dd}\\,{{H}_{2}}S{{O}_{4}}}}=\\dfrac{49.100\\%}{9,8\\%}=500\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1842},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" 0,5mol $CuO$ t\u00e1c d\u1ee5ng v\u1eeba \u0111\u1ee7 v\u1edbi:","select":["A. 0,5mol $H_2SO_4.$ ","B. 0,25mol $HCl.$","C. 0,5mol $HCl.$ ","D. 0,1mol $H_2SO_4$"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned} & CuO+{{H}_{2}}S{{O}_{4}}\\to CuS{{O}_{4}}+{{H}_{2}}O \\\\ & CuO+2HCl\\to CuC{{l}_{2}}+{{H}_{2}}O \\\\ & {{n}_{CuO}}={{n}_{{{H}_{2}}S{{O}_{4}}}}=\\dfrac{1}{2}{{n}_{HCl}} \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1843},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" H\u00f2a tan h\u1ebft $6,2$ gam $Na_2O$ v\u00e0o $H_2O$ \u0111\u01b0\u1ee3c dung d\u1ecbch $X$. Th\u1ec3 t\u00edch dung d\u1ecbch $HCl$ $0,5M$ c\u1ea7n \u0111\u1ec3 ph\u1ea3n \u1ee9ng h\u1ebft v\u1edbi dung d\u1ecbch $X$ l\u00e0:","select":["A. 100 ml ","B. 200 ml","C. 300 ml ","D. 400 ml"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{N{{a}_{2}}O}}=\\dfrac{6,2}{62}=0,1\\,mol \\\\ & N{{a}_{2}}O+{{H}_{2}}O\\to 2NaOH \\\\ & NaOH+HCl\\to NaCl+{{H}_{2}}O \\\\ & {{n}_{HCl}}={{n}_{NaOH}}=2{{n}_{N{{a}_{2}}O}}=0,2\\,mol \\\\ & V_{HCl}=\\dfrac{0,2}{0,5}=0,4=400\\,ml \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":1844},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" H\u00f2a tan h\u1ebft $12,4$ gam Natrioxit v\u00e0o n\u01b0\u1edbc thu \u0111\u01b0\u1ee3c $500ml$ dung d\u1ecbch $A$ . N\u1ed3ng \u0111\u1ed9 mol c\u1ee7a dung d\u1ecbch $A$ l\u00e0 :","select":["A. 0,8M ","B. 0,6M","C. 0,4M ","D. 0,2M"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{N{{a}_{2}}O}}=\\dfrac{12,4}{62}=0,2\\,mol \\\\ & N{{a}_{2}}O+{{H}_{2}}O\\to 2NaOH \\\\ & {{n}_{NaOH}}={{2n}_{N{{a}_{2}}O}}=0,4\\,mol \\\\ & {{C}_{M\\,(NaOH)}}=\\dfrac{0,4}{0,5}=0,8\\,M \\\\ \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1845},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" \u0110\u1ed1t ch\u00e1y 16,8 gam s\u1eaft trong kh\u00ed \u00f4xi \u1edf nhi\u1ec7t \u0111\u1ed9 cao thu \u0111\u01b0\u1ee3c 16,8 gam $Fe_3O_4$. Hi\u1ec7u su\u1ea5t ph\u1ea3n \u1ee9ng l\u00e0:","select":["A. 71,4% ","B. 72,4%.","C. 73,4% ","D. 74,4%."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{Fe}}=0,3\\,mol \\\\ & 3Fe+2{{O}_{2}}\\xrightarrow{{{t}^{0}}}F{{e}_{3}}{{O}_{4}} \\\\ & {{n}_{F{{e}_{3}}{{O}_{4}}}}=\\dfrac{1}{3}{{n}_{Fe}}=0,1\\,mol \\\\ & {{m}_{F{{e}_{3}}{{O}_{4}}}}=0,1.232=23,2\\,gam \\\\ & H=\\dfrac{16,8}{23,2}.100\\%\\approx 72,4\\% \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1846},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":" Nh\u00fang \u0111inh s\u1eaft v\u00e0o dung d\u1ecbch $CuSO_4$, khi l\u1ea5y \u0111inh s\u1eaft ra kh\u1ed1i l\u01b0\u1ee3ng t\u0103ng 0,2g so v\u1edbi ban \u0111\u1ea7u. Kh\u1ed1i l\u01b0\u1ee3ng kim lo\u1ea1i \u0111\u1ed3ng b\u00e1m v\u00e0o s\u1eaft:","select":["A. 0,2 g ","B. 1,6 g.","C. 3,2 g. ","D. 6,4 g."],"hint":"","explain":"<span class='basic_left'>G\u1ecdi s\u1ed1 mol Fe ph\u1ea3n \u1ee9ng l\u00e0 x (mol). <br\/>$\\begin{aligned}& Fe+CuS{{O}_{4}}\\to FeS{{O}_{4}}+Cu \\\\ & x\\,mol\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,x\\,mol \\\\ \\end{aligned}$ <br\/>Kh\u1ed1i l\u01b0\u1ee3ng \u0111inh s\u1eaft t\u0103ng l\u00ean b\u1eb1ng hi\u1ec7u c\u1ee7a kh\u1ed1i l\u01b0\u1ee3ng Cu t\u1ea1o ra v\u00e0 kh\u1ed1i l\u01b0\u1ee3ng Fe \u0111\u00e3 ph\u1ea3n \u1ee9ng.<br\/> Ta c\u00f3 mCu \u2013 mFe = 0,2g<br\/>$\\begin{aligned}& =>64x - 56x = 0,2 \\\\ & =>\\,8x=0,2 \\\\ & =>\\,x=0,025 mol \\\\ &\\end{aligned}$ <br\/>Kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a Cu t\u1ea1o ra l\u00e0: <br\/>$\\begin{aligned}&{{m}_{Cu}}=0,025.64=1,6\\,gam \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span><\/span> ","column":2}]}],"id_ques":1847},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" Kh\u1eed 16 gam $Fe_2O_3$ b\u1eb1ng $CO$ d\u01b0 , s\u1ea3n ph\u1ea9m kh\u00ed thu \u0111\u01b0\u1ee3c cho \u0111i v\u00e0o dung d\u1ecbch $Ca(OH)_2$ d\u01b0 thu \u0111\u01b0\u1ee3c a gam k\u1ebft t\u1ee7a. Gi\u00e1 tr\u1ecb c\u1ee7a a l\u00e0 ","select":["A. 10 g ","B. 20 g","C. 30 g ","D. 40 g"],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& {{n}_{F{{e}_{2}}{{O}_{3}}}}=0,1\\,mol \\\\ & F{{e}_{2}}{{O}_{3}}+3CO\\xrightarrow{{{t}^{0}}}2Fe+3C{{O}_{2}} \\\\ & C{{O}_{2}}+Ca{{(OH)}_{2}}\\to CaC{{O}_{3}}\\downarrow +{{H}_{2}}O \\\\ & {{n}_{\\downarrow }}={{n}_{C{{O}_{2}}}}=3{{n}_{F{{e}_{2}}{{O}_{3}}}}=0,3\\,mol \\\\ & {{m}_{\\downarrow }}=0,3.100=30\\,gam \\end{aligned}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 C.<\/span><\/span> ","column":2}]}],"id_ques":1848},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":" H\u00f2a tan ho\u00e0n to\u00e0n 12,1 gam h\u1ed7n h\u1ee3p b\u1ed9t CuO v\u00e0 ZnO c\u1ea7n 100 ml dung d\u1ecbch HCl 3M. Th\u00e0nh ph\u1ea7n ph\u1ea7n tr\u0103m theo kh\u1ed1i l\u01b0\u1ee3ng hai oxit tr\u00ean l\u1ea7n l\u01b0\u1ee3t l\u00e0:","select":["A. 33,06% v\u00e0 66,94% ","B. 66,94% v\u00e0 33,06%","C. 33,47% v\u00e0 66,53% ","D. 66,53% v\u00e0 33,47%"],"hint":"","explain":"<span class='basic_left'>\u0110\u1eb7t x, y l\u1ea7n l\u01b0\u1ee3t l\u00e0 s\u1ed1 mol c\u1ee7a CuO, ZnO trong h\u1ed7n h\u1ee3p <br\/>Kh\u1ed1i l\u01b0\u1ee3ng h\u1ed7n h\u1ee3p = 12,1 gam n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh: <br\/>80x + 81y = 12,1 (1) <br\/>$\\begin{aligned}& CuO+2HCl\\to CuC{{l}_{2}}+{{H}_{2}}O \\\\ & ZnO+2HCl\\to ZnC{{l}_{2}}+{{H}_{2}}O \\\\ & {{n}_{HCl}}=0,3\\,mol \\\\ & =>\\,2x\\,+2y=0,3(2) \\\\ \\end{aligned}$ <br\/>Gi\u1ea3i (1) v\u00e0 (2) ta \u0111\u01b0\u1ee3c x = 0,05 mol; y = 0,1 mol <br\/>$\\begin{aligned}& \\%{{m}_{CuO}}=\\dfrac{0,05.80}{12,1}.100\\%\\approx 33,06\\% \\\\ & \\%{{m}_{ZnO}}=100\\%-33,06\\%=66,94\\% \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><\/span> ","column":2}]}],"id_ques":1849},{"time":24,"part":[{"title":"Ch\u1ecdn ph\u01b0\u01a1ng \u00e1n \u0110\u00daNG","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" Th\u00eam 20 g HCl v\u00e0o 480 gam dung d\u1ecbch HCl 5%, thu \u0111\u01b0\u1ee3c dung d\u1ecbch m\u1edbi c\u00f3 n\u1ed3ng \u0111\u1ed9:","select":["A. 9,8% ","B. 8,7%.","C. 8,9%. ","D. 8,8%."],"hint":"","explain":"<span class='basic_left'>$\\begin{aligned}& \\text{Tng kh }\\!\\!\\grave{\\mathrm{e}}\\!\\!\\text{ i l }\\!\\!\\hat{\\mathrm{i}}\\!\\!\\text{ ng c }\\!\\!\\tilde{\\mathrm{n}}\\!\\!\\text{ a HCl l }\\!\\!\\mu\\!\\!\\text{ :} \\\\ & \\text{20 + }\\dfrac{480.5\\%}{100\\%}=44\\,gam \\\\ & {{m}_{dd}}=20+480=500\\,gam \\\\ & C{{\\%}_{HCl}}=\\dfrac{44}{500}.100\\%=8,8\\% \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D.<\/span><\/span> ","column":2}]}],"id_ques":1850}],"lesson":{"save":0,"level":2}}

Điểm của bạn.

Câu hỏi này theo dạng chọn đáp án đúng, sau khi đọc xong câu hỏi, bạn bấm vào một trong số các đáp án mà chương trình đưa ra bên dưới, sau đó bấm vào nút gửi để kiểm tra đáp án và sẵn sàng chuyển sang câu hỏi kế tiếp

Trả lời đúng trong khoảng thời gian quy định bạn sẽ được + số điểm như sau:
Trong khoảng 1 phút đầu tiên + 5 điểm
Trong khoảng 1 phút -> 2 phút + 4 điểm
Trong khoảng 2 phút -> 3 phút + 3 điểm
Trong khoảng 3 phút -> 4 phút + 2 điểm
Trong khoảng 4 phút -> 5 phút + 1 điểm

Quá 5 phút: không được cộng điểm

Tổng thời gian làm mỗi câu (không giới hạn)

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Bấm vào đây nếu phát hiện có lỗi hoặc muốn gửi góp ý